F1 = 9x10^9 ( ((2x10^-6)(-8x10^-6))/ ( 0.1 )^2 ) = -14.4N
F2 = 9x10^9 ( ((3x10^-6)(-8x10^-6))/ ( 0.1 )^2 ) = -21.6N
F = √ (F1^2+F2^2+2(F1)(F2)(cos(θ)))
F = √ ((14.4)^2 + (21.6)^2 + 2(14.4)(21.6)(Cos(60)))
F = 31.38N
ArcTan((14.4Sen(60))/(21.6+14.4Cos(60)) = 23.41º
Charges of +2.0, +3.0 and -8.0µC are placed at the vertices of an equilateral triangle of a side 10 cm. Calculate the magnitude of the force acting on the -8µC charge due to the other two loads.
2 answers
Case 2
F1 = 9x10^9 ( ((2x10^-6)(8x10^-6))/ ( 0.1 )^2 ) = 14.4N
F2 = 9x10^9 ( ((3x10^-6)(8x10^-6))/ ( 0.1 )^2 ) = 21.6N
Fex = 14.4Cos(120) - 21.6Cos(0) = -28.8
Fey = 14.4Sen(120) + 21.6Sen(0) = 12.47076581
Fe = √ ((28.8)^2+(12.47076581)^2) = 31.38407239N
ArcTan((14.4)/(21.6)) = 33.69006753
180 - 33.69006753 = 146.3099325º
F1 = 9x10^9 ( ((2x10^-6)(8x10^-6))/ ( 0.1 )^2 ) = 14.4N
F2 = 9x10^9 ( ((3x10^-6)(8x10^-6))/ ( 0.1 )^2 ) = 21.6N
Fex = 14.4Cos(120) - 21.6Cos(0) = -28.8
Fey = 14.4Sen(120) + 21.6Sen(0) = 12.47076581
Fe = √ ((28.8)^2+(12.47076581)^2) = 31.38407239N
ArcTan((14.4)/(21.6)) = 33.69006753
180 - 33.69006753 = 146.3099325º
2 answers